Notes to Wolfgang Spohn’s “The Laws of Belief”

Claus-Peter Wirth

FB AI, Hochschule Harz, D–38855 Wernigerode, Germany

   wirth@logic.at

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Notes to Wolfgang Spohn’s “The Laws of Belief”

Claus-Peter Wirth FB AI, Hochschule Harz, D–38855 Wernigerode, Germany

                        wirth@logic.at

March 14, 2014

Abstract

This are some private notes to [Spohn, 2012].

Keywords: Ranking Semantics

1 (Negative) Ranking Functions

The set W in the following definition is meant to be a set of mutually exclusive cases, typically formed by a conjunction of one concrete value for each observed variable, such as

temp = 225.5 K ∧ uptime = 503 s

or emu(edna) = true ∧ bird(edna) = true ∧ flies(edna) = false.

A subset of W is called a proposition and is meant to be the case that consists in any of the cases of its elements, i.e. the disjunction of its elements. The name “proposition” for a subset of W is justified insofar as it is meant to be something similar to a full disjunctive normal form.

It is assumed that W ≠ ∅, so that ∅ can represent FALSE, and W can represent TRUE.

Definition 1.1 (Algebra) A is an [complete] algebra over W if we have W ≠ ∅, A is a subset of the power set of W, and the following holds:

1. W ∈ A.

2. A, B ∈ A implies A ∪ B ∈ A. [Moreover, for each B ⊆ A, we have ⋃ B ∈ A.]

3. A ∈ A implies \overline{A} ∈ A, where “ \overline{A} ” denotes A \ A.

Using Definition 1.1(3) and \overline{\overline{A} ∪ \overline{B}} = A ∩ B, &c., we get:

Corollary 1.2 Let A be an [complete] algebra over W. Then:

1. ∅ ∈ A.

2. A, B ∈ A implies A ∩ B ∈ A. [Moreover, for each B ⊆ A, we have ⋂ B ∈ A.]

Let “ω” denote the first infinite ordinal number and let ω+1 be its successor ω ∪ {ω}. Let “inf” denote the infimum over subsets of ω+1, i.e. inf A = min A for A ⊆ ω+1 with A ≠ ∅, and inf ∅ = ω. Moreover, let us define, for any n ≺ ω : n + ω := ω and ω − n := ω, and ω − ω := 0.

Definition 1.3 (Set-Wise (Negative) Ranking Function) Let A be an algebra over W. A set-wise (negative) ranking function for A is a function κ : A → ω+1 with

1. κ(W) = 0. “Global disjunction is believed.”

2. κ(⋃ B) = inf { κ(B) | B ∈ B } for each B ⊆ A with ⋃ B ∈ A. “Law of Disjunction”

Let A|A denote the set of words of length 3 consisting of a proposition from A, the sign “|” and a further proposition from A. We will read B|A as “B under condition of A”. We extend any set-wise ranking function κ for A from A to A|A by (A, B ∈ A):

κ(B|A) := \{ \begin{matrix}ω & if A∩B = ∅ and κ(A) = ω \\ κ(A∩B) − κ(A) & otherwise \end{matrix} \} .

Note that — to make clear that a higher rank means a higher level of disbelief instead of belief — the defined ranking functions are sometimes called “negative” ranking functions.^{1} For instance, the Law of Disjunction says that our disbelief in a disjunction of propositions is equal to our smallest disbelief in any of the propositions.

Taking B in Definition 1.3(2) to be one of ∅, \{B, \overline{B} \}, \{B, C ∩ \overline{B} \}, we get the following three items, respectively:

Corollary 1.4 Let A be an algebra over W. Let κ be a set-wise ranking function for A. Then:

1. κ(∅) = ω. “Empty disjunction is not believed.”

2. κ(B) ≠ 0 implies κ( \overline{B} ) = 0 for each B ∈ A. “Law of Negation”

3. For B, C ∈ A with B ⊆ C, we have κ(B) ⪰ κ(C).

Note that the Law of Negation says that if we disbelieve a proposition then we have to believe the complementary proposition.

1This negativeness is in opposition to standard probability theory and also to positive and two-sided ranking functions, cf. [Spohn, 2012, § 5.2, pp. 69–77]. Positive and two-sided ranking functions do not provide any extra in formation, but their theory is typically more awkward w.r.t. the arithmetic expressions.

Lemma 1.5 Let A be an algebra over W. Let κ be a set-wise ranking function for A. For A, B, C ∈ A:

1. κ(B|A) = κ(A∩B) − κ(A)

if A∩B ≠ ∅ or κ(A) ≠ ω.

2. κ(A∩B) = κ(A) + κ(B|A)

if A∩B ≠ ∅ or κ(A) ≠ ω. “Law of Conjunction”

3. κ(B|A) = ω iff A∩B = ∅ or κ(A) ≺ κ(A∩B) = ω.

4. κ_{A}(C) := κ(C|A) (C ∈ A) defines κ_{A} as a set-wise ranking function forA unless A = ∅.

5. κ(B|A) = (κ(B) + κ(A|B)) − κ(A)

if A∩B ≠ ∅ or κ(A) ≠ ω ≠ κ(B). “Bayes’ Theorem^{2}”

6. κ(B) = inf { κ(A) + κ(B|A) | A ∈ P ∧ A∩B ≠ ∅ } if P ⊆ A with ⋃ P = W. “Formula of the Total Rank”

7. Assume A∩C ≠ ∅ ∨ κ(A) ≠ ω, B∩C ≠ ∅ ∨ κ(B) ≠ ω, and κ(C|A) ⪯ κ(C|B). Then:

κ(C|A) ⪯ κ(C|A∪B) ⪯ κ(C|B). “Law of Disjunctive Conditions”

More precisely, we have:

(a) If κ(A) ⪯ κ(B), then κ(A∩C) ⪯ κ(B∩C).

(b) κ(C|A) = κ(C|A∪B) in case of κ(A) ⪯ κ(B).

(c) κ(C|A) ≺ κ(C|A∪B) in case of κ(A) ≻ κ(B) and κ(A∩C) ⪯ κ(A∩B).

(d) If κ(A∩C) ⪰ κ(B∩C), then κ(A) ⪰ κ(B).

(e) κ(C|A∪B) = κ(C|B) in case of κ(A∩C) ⪰ κ(B∩C).

(f) κ(C|A∪B) ≺ κ(C|B) in case of κ(A∩C) ≺ κ(B∩C) and κ(A) ⪰ κ(B).

This means that — in the irreflexive case κ(C|A) ≺ κ(C|B) — we have:

(g) κ(C|A) ≺ κ(C|A∪B) iff κ(A) ≻ κ(B).

(h) κ(C|A∪B) ≺ κ(C|B) iff κ(A∩C) ≺ κ(B∩C).

P (B|A) = \frac{P (B) ∗ P (A|B)}{P (A)} .

2This is called “Bayes’ Theorem” after the English philosopher Thomas Bayes (1701–1761), who formulated a restricted version of what is called “Bayes’ Theorem” in probability theory today:

In general, theorems in probability theory strongly suggest theorems in ranking theory as describe as follows in [Spohn, 2012, p. 82]:

“adding probabilities translates into taking the minimum of negative ranks, multiplication of

probabilities into addition of negative ranks, and division of probabilities into subtraction of

negative ranks.”

Example 1.6 Set W := {0, 1, 2}, A := �(W).

1. Set κ({0}) = 0, κ({1}) = 1, κ({2}) = 2. Set A := {1}, B := {2}. Then, by Corollary 1.4, κ(A∩B) = κ(∅) = ω. Moreover, by Lemma 1.5(1), we have κ(B|A) = κ(A∩B) − κ(A) = ω − 1 = ω. As the second option of the condition of Lemma 1.5(2) is given, we indeed have κ(A∩B) = κ(A) + κ(B|A), namely ω = 1 + ω. But as ω ≠ ω + 1, we do not have κ(A∩B) = κ(B|A) + κ(A).

This means that the two arguments on the right-hand side of Lemma 1.5(2) cannot be commuted in case of κ(A) ≠ ω.

2. Set κ({0}) := 0, κ({1}) := ω. Set A := B := {1}. Then A∩B = A = B ≠ ∅. Moreover, κ(A|B) = κ(B|A) = 0 by Lemma 1.5(1). Thus κ(B|A) ≠ ω = ω + 0 = ω + (0 − ω) = κ(B) + (κ(A|B) − κ(A)).

This means that the parentheses on the right-hand side of Lemma 1.5(5) cannot be dropped in case of A∩B ≠ ∅.

Proof of Lemma 1.5

\underline{\underline{\underline{(1):}}} By Definition 1.3.

\underline{\underline{\underline{(2):}}} On the one hand, if κ(A) ≠ ω, then we have κ(B|A) = κ(A∩B) − κ(A) by (1),

and then, in case of κ(A∩B) ≠ ω, both κ(A∩B) = κ(A) + κ(B|A) and κ(A∩B) = κ(B|A)+κ(A); but the former even holds in case of κ(A∩B) = ω because then κ(B|A) = ω. If, on the other hand, κ(A) = ω, then we have κ(A∩B) = ω according to Corollary 1.4; moreover, we have κ(B|A) = κ(A∩B) − κ(A) = ω − ω = 0 by Definition 1.3 and the restriction of (2). We then have κ(A∩B) = ω = ω + 0 = κ(A) + κ(B|A) again.

\underline{\underline{\underline{(3):}}} Assume A∩B = ∅. Then, in case of κ(A) = ω, we have κ(B|A) = ω directly by

Definition 1.3. Otherwise, in case of κ(A) ≠ ω, by (1) and Corollary 1.4(1) we get κ(B|A) = κ(A∩B) − κ(A) = κ(∅) − κ(A) = ω − κ(A) = ω again.

\underline{\underline{\underline{(4):}}} Assume A ≠ ∅. Because of A∩W = A ≠ 0, we have

κ_{A}(W) = κ(W |A) = κ(A∩W) − κ(A) = κ(A) − κ(A) = 0. So the global disjunction is disbelieved indeed. Moreover, for showing the Law of Disjunction, suppose that B ⊆ A with ⋃ B ∈ A. \underline{(a) A∩ ⋃ B = ∅ and κ(A) = ω:} Then: (α) For each B ∈ B, we have A∩B = ∅.

κ_{A}(⋃ B) = κ(⋃ B|A) =^{(a)} ω = inf { ω |B ∈ B } =^{(α)} inf { κ(B|A)|B ∈ B } = inf { κ_{A}(B)|B ∈ B }.

\underline{(b) A∩ ⋃ B ≠ ∅ or κ(A) ≠ ω:} Set B′ := \{ B | B ∈ B ∧ A∩B ≠ ∅ \}.

Set B′′ := { A∩B | B ∈ B }. Then:

(β) For each B′ ∈ B′, we have A∩B′ ≠ ∅. (γ) If κ(A) = ω, then B′ ≠ ∅. (δ) According to (3), for B ∈ B\B′, we have κ(B|A) = ω. (ε) B′′ ⊆ A and ⋃ B′′ = A ∩ ⋃ B′ = A ∩ ⋃ B ∈ A. (ζ) From (ε), by Definition 1.3(2), we get: κ(⋃ B′′) = inf { κ(B′′) | B′′ ∈ B′′ }.

κ_{A}(⋃ B) = κ(⋃ B|A) =^{(b)} κ(A ∩ ⋃ B) − κ(A) =^{(ε)} κ(⋃ B′′) − κ(A)
=^{(ζ)} inf { κ(B′′) | B′′ ∈ B′′ } − κ(A) = inf { κ(A∩B′) | B′ ∈ B′ } − κ(A)
=^{(γ)} inf { κ(A∩B′) − κ(A) | B′ ∈ B′ } =^{(β)} inf { κ(B′|A) | B′ ∈ B′ }
= inf { inf { κ(B′|A) | B′ ∈ B′ }, ω }
=^{(δ)} inf { inf { κ(B′|A) | B′ ∈ B′ }, inf { κ(B|A) | B ∈ B\B′ } }
= inf { κ(B|A) | B ∈ B } = inf { κ_{A}(B) | B ∈ B }.

\underline{\underline{\underline{(5):}}} If A∩B ≠ ∅, then we have κ(B|A) =^{(1)} κ(A ∩ B) − κ(A) = κ(B ∩ A) − κ(A) =^{(2)}

(κ(B) + κ(A|B)) − κ(A). Otherwise, in case of A∩B = ∅ and κ(A) ≠ ω ≠ κ(B),

we have κ(B|A) =^{(3)} ω = κ(B) + ω − κ(A) =^{(3)} κ(B) + κ(A|B) − κ(A).

\underline{\underline{\underline{(6):}}} Set P′ := \{ A ∈ P | A∩B ≠ ∅ \}. Set P′′ := \{ A′∩B | A′ ∈ P′ \}. Then: (α) For each A′ ∈ P′, we have A′∩B ≠ ∅, and then, by (2): κ(A′∩B) = κ(A′)+κ(B|A′). (β) P′′ ⊆ A and ⋃ P′′ = B ∩ ⋃ P′ = B ∩ ⋃ P = B ∩ W = B ∈ A. (γ) From (β), by Definition 1.3(2), we get: κ(⋃ P′′) = inf { κ(A′′) | A′′ ∈ P′′ }.

κ_{A}(B) =^{(β)} κ(⋃ P′′) =^{(γ)} inf { κ(A′′) | A′′ ∈ P′′ } = inf { κ(A′∩B) | A′ ∈ P′ }
=^{(α)} inf { κ(A′) + κ(B|A′) | A′ ∈ P′ }
= inf { κ(A) + κ(B|A) | A ∈ P ∧ A∩B ≠ ∅ }

\underline{\underline{\underline{(7):}}} Let us assume Assume A∩C ≠ ∅ ∨ κ(A) ≠ ω and B∩C ≠ ∅ ∨ κ(B) ≠ ω.

Then, by Definition 1.3(2), we also have (A∪B)∩C ≠ ∅ or κ(A∪B) ≠ ω. Then, by (1) and Definition 1.3(2) we have:

(α) κ(C|A) = κ(A∩C) − κ(A). (β) κ(C|B) = κ(B∩C) − κ(B). (γ) κ(C|A∪B) = κ((A∪B)∩C) − κ(A∪B) = κ((A∩C)∪(B∩C)) − κ(A∪B)

= inf{κ(A∩C), κ(B∩C)} − inf{κ(A), κ(B)}.

Let us assume according to the assumption in the lemma:

(δ) κ(C|A) ⪯ κ(C|B).

By (α, β), we get from (δ):

(ε) κ(A∩C) − κ(A) ⪯ κ(B∩C) − κ(B).

\underline{\underline{(a), (d):}} For both items it suffices to refute (κ(A) ⪯ κ(B)) ∧ (κ(B∩C) ⪯ κ(A∩C))

and the additional assumption that one of the two steps is actually irreflexive

(i.e. a step via ≺). In this case, however, we get κ(A∩C) − κ(A) ⪰ κ(B∩C) − κ(A) ⪰ κ(B∩C) − κ(B). Under the additional assumption one of these two steps is irreflexive and so we get κ(A∩C) − κ(A) ≻ κ(B∩C) − κ(B), which contradicts (ε).

\underline{\underline{(b):}} Let us assume κ(A) ⪯ κ(B). By (a) we get κ(A∩C) ⪯ κ(B∩C). Then:

κ(C|A) =^{(α)} κ(A∩C)−κ(A) = inf{κ(A∩C), κ(B∩C)}−inf{κ(A), κ(B)} =^{(γ)} κ(C|A∪B).

\underline{\underline{(e):}} Let us assume κ(A∩C) ⪰ κ(B∩C). By (d) we get κ(A) ⪰ κ(B). Then:

κ(C|B) =^{(α)} κ(B∩C)−κ(B) = inf{κ(A∩C), κ(B∩C)}−inf{κ(A), κ(B)} =^{(γ)} κ(C|A∪B).

\underline{\underline{κ(A) ⪰ κ(B) and κ(A∩C) ⪯ κ(B∩C) :}} Then κ(C|A∪B) =^{(γ)} inf\{κ(A∩C), κ(B∩C)\} −

inf{κ(A), κ(B)} = κ(A∩C) − κ(B), i.e.:

(ζ) κ(C|A∪B) = κ(A∩C) − κ(B). (η) Moreover, κ(B) ≠ ω.

\underline{(c):} Let us assume that the first element of our case assumption is an irreflexive step.

Then: κ(C|A) =^{(α)} κ(A∩C) − κ(A) ≺ κ(A∩C) − κ(B) =^{(ζ)} κ(C|A∪B).

\underline{(f ):} Let us assume that the second element of our case assumption is an irreflexive step.

Then: κ(C|A∪B) =^{(ζ)} κ(A∩C) − κ(B) ≺^{(η)} κ(B∩C) − κ(B) =^{(β)} κ(C|B).

\underline{\underline{“Law of Disjunctive Conditions”:}} This holds for κ(A) ⪯ κ(B) by (b), and for

κ(A∩C) ⪰ κ(B∩C) by (e). In the remaining case of

(κ(A) ≻ κ(B)) ∧ (κ(A∩C) ≺ κ(B∩C)), it holds by (c, f ).

\underline{\underline{(g):}} \underline{“⇒”:} By (b), the negation of κ(A) ≻ κ(B) implies the negation of κ(C|A) ≺ κ(C|A∪B).

\underline{“⇐”:} In case of κ(A∩C) ⪯ κ(B∩C), this holds by (c). Otherwise, it holds by (e): κ(C|A) ≺ κ(C|B) = κ(C|A∩B).

\underline{\underline{(h):}} \underline{“⇒”:} By (e), the negation of κ(A∩C) ≺ κ(B∩C) implies the negation of κ(C|A) ≺ κ(C|A∪B).

\underline{“⇐”:} In case of κ(A) ⪰ κ(B), this holds by (f ). Otherwise, it holds by (b): κ(C|A∩B) = κ(C|A) ≺ κ(C|B). Q.e.d. (Lemma 1.5)

Definition 1.7 (Point-Wise (Negative) Ranking Function) Let A be an algebra over W.

1. A point-wise (negative) ranking function for A is a function κ : W → ω+1

with κ⟨{0}⟩ ≠ ∅, i.e. there is a w ∈ W with κ(w) = 0. ^{3} “Something is believed.”

2. The set-wise ranking function κ̂ of a point-wise ranking function κ for A is defined by κ̂(B) = inf { κ(w) | w ∈ B } (B ∈ A).

3. Let W |A denote the set of words of length 3 consisting of an element from W, the sign “|” and a proposition from A. We will read w|A as “w under condition of A”. We extend any point-wise ranking function κ for A from W to W |A by (w ∈ W, A ∈ A):

κ(w|A) := \{ \begin{matrix}ω & if w ∉ A \\ κ(w) − κ̂(A) & if w ∈ A \end{matrix} \} .

4. Let W |A↑ω+1 denote the set of words of length 5 consisting of an element from W, the sign “|”, a proposition from A, the sign “↑”, and an element from ω+1. We will read w|A↑α as “w under condition of A with evidence α”. We extend any point-wise ranking function κ for A from W to W |A↑ω+1 by (w ∈ W, A ∈ A, α ∈ ω+1):

κ(w|A↑α) := \{ \begin{matrix}ω & if w ∉ A and α = ω \\ (α − inf\{κ̂(A), α\}) + κ(w) & if w ∉ A and α ≠ ω \\ κ(w) − inf\{κ̂(A), α\} & if w ∈ A \end{matrix} \}.

Corollary 1.8 Let A be an algebra over W. Let κ be a point-wise ranking function for A.

1. Let κ̂ be the set-wise ranking function of κ. Then: For w ∈ A ∈ A, we have κ(w) ⪰ κ̂(A).

2. For w ∈ W, A ∈ A :

κ(w|A↑0)  = κ(w),
κ(w|A↑α)  = \{ \begin{matrix}κ(w) & if w ∉ A \\ κ(w) − α & if w ∈ A \end{matrix} \}  if α ≺ κ̂(A),
κ(w|A↑κ̂(A)+β) = \{ \begin{matrix}β + κ(w) & if w ∉ A \\ κ(w) − κ̂(A) & if w ∈ A \end{matrix} \} if β ≺ ω,
κ(w|A↑ω)  = κ(w|A).

Definition 1.9 (Point-Wise Ranking Function of Set-Wise Ranking Function) Let A be a complete algebra over W. The point-wise ranking function κ̇ of a set-wise ranking function κ for A is defined by κ̇(w) = κ(⋂ { B ∈ A | w ∈ B }) (w ∈ W).

3In his related definition, [Spohn, 2012, Definition 5.5, p. 70] additionally requires κ⟨{α}⟩ ∈ A for each α ∈ ω+1.

Lemma 1.10 Let A be a complete algebra over W. Let κ be a set-wise ranking function for A. Then:

1. κ̇ (the point-wise ranking function of κ) is indeed a point-wise ranking function for A.

2. ˆ κ̇ (the set-wise ranking function of κ̇) satisfies ˆ κ̇ = κ.

Proof of Lemma 1.10

\underline{(1):} By Corollary 1.2(2), κ̇ : W → ω+1 is well-defined because A is a complete alge-

bra over W. Moreover, by Definition 1.3(2,1), we have inf { κ̇(w) | w ∈ W } = inf { κ(⋂ { B ∈ A | w ∈ B }) | w ∈ W } = κ(⋃ { ⋂ { B ∈ A | w ∈ B } | w ∈ W }) = κ(W) = 0. Thus, there must be some w ∈ W with κ̇(w) = 0, i.e. κ̇⟨{0}⟩ ≠ ∅.

\underline{(2):} By Definitions 1.7 and 1.9 and Definition 1.3(2), we have for every B′ ∈ A:

ˆκ̇(B′) = inf { κ̇(w) | w ∈ B′ } = inf { κ(⋂ { B ∈ A | w ∈ B }) | w ∈ B′ } = κ(⋃ { ⋂ { B ∈ A | w ∈ B } | w ∈ B′ }) = κ(B′). Indeed, we have B′ = ⋃ { ⋂ { B ∈ A | w ∈ B } | w ∈ B′ }, because, for each w ∈ B′, we have w ∈ ⋂ { B ∈ A | w ∈ B } ⊆ B′. Q.e.d. (Lemma 1.10)

Lemma 1.11 Let A be an algebra over W. Let κ be a point-wise ranking function for A. Then:

1. κ̂ (the set-wise ranking function of κ) is indeed a set-wise ranking function for A.

2. ˙ κ̂ (the set-wise ranking function of κ̂) satisfies ˙ κ̂(w′) = κ(w′) for each w′ ∈ W with inf { κ(w) | w ∈ ⋂ { B ∈ A | w′ ∈ B } } = κ(w′), which is the case e.g. if {w′} ∈ A or if κ⟨{κ(w′)}⟩ ∈ A.

The following example shows that the restriction in Lemma 1.11(2) is indeed necessary:

Example 1.12 Set W := {0, 1}, A := {∅, W }, κ(0) := 0, κ(1) := 1. Then κ̂(∅) = ω, κ̂(W) = 0. Then ˙ κ̂(0) = ˙ κ̂(1) = 0, i.e. ˙ κ̂ ≠ κ.

Proof of Lemma 1.11

\underline{(1):} Because of κ⟨\{0\}⟩ ≠ ∅, by Definition 1.7, we have κ̂(W) = inf \{ κ(w) | w ∈ W \} =

inf ({ κ(w) | w ∈ W } ∪ {0}) = 0. Moreover, for B ⊆ A with ⋃ B ∈ A, we have κ̂(⋃ B) = inf { κ(w) | w ∈ ⋃ B } = inf { inf { κ(w) | w ∈ B } | B ∈ B } = inf { κ̂(B) | B ∈ B }.

\underline{(2):} By Definitions 1.7 and 1.9, we have for every w′ ∈ W with

inf { κ(w) | w ∈ ⋂ { B ∈ A | w′ ∈ B } } = κ(w′): ˙κ̂(w′) = κ̂(⋂ { B ∈ A | w′ ∈ B }) = inf { κ(w) | w ∈ ⋂ { B ∈ A | w′ ∈ B } } = κ(w′).

Q.e.d. (Lemma 1.11)

Lemma 1.13 Let A be an algebra over W. Let κ be a point-wise ranking function for A. For w ∈ W, A, B ∈ A, α, β ∈ ω+1 :

1. κ(w|A) = κ(w) − κ̂(A) unless w ∉ A.

2. κ(w) = κ̂(A) + κ(w|A) unless w ∉ A. “Law of Conjunction”

3. κ(w|A) = ω iff w ∉ A or κ̂(A) ≺ κ(w) = ω.

4. κ_{A}(v) := κ(v|A) (v ∈ W) defines κ_{A} as a point-wise ranking function forAunless A = ∅.

5. κ_{A↑α}(v) := κ(v|A↑α) (v ∈ W) defines κ_{A↑α} as a point-wise ranking function for A unless A = ∅.

6. ??? Assume κ_{A↑α}, κ_{B↑β}, (κ_{A↑α})_{B↑β}, and (κ_{B↑β})_{A↑α} to be given as in (5) for A ≠ ∅ ≠ B. Then (κ_{A↑α})_{B↑β} = (κ_{B↑β})_{A↑α}. ???

Proof of Lemma 1.13

\underline{\underline{\underline{(1):}}} By Definition 1.7(3).

\underline{\underline{\underline{(2):}}} Let us assume w ∈ A. Then we have κ(w|A) = κ(w)−κ̂(A) by (1), and then, in case

of κ(w) ≺ ω, we have κ̂(A) ⪯ κ(w) ≺ ω by Corollary 1.8, and so κ(w|A) ≺ ω, which means that we have both κ(w) = κ̂(A) + κ(w|A) and κ(w) = κ(w|A) + κ̂(A). Both also hold in case of κ(w) = ω and κ̂(A) = ω because then κ(w|A) = 0. In the remaining case of κ(w) = ω and κ̂(A) ≺ ω, we get κ(w|A) = ω, and so the former of the two equations holds even in this case.

\underline{\underline{\underline{(3):}}} By Definition 1.7(3) and the same case analysis as in (2).

\underline{\underline{\underline{(4):}}} Let us assume A ≠ ∅. By A ≠ ∅ and by Definition 1.7(2), there must be some

w ∈ A with κ(w) = κ̂(A). Then, by Definition 1.7(3), we have κ_{A}(w) = κ(w|A) = κ(w) − κ̂(A) = κ̂(A) − κ̂(A) = 0.

\underline{\underline{\underline{(5):}}} Let us assume A ≠ ∅. We have to do a case analysis according to Corollary 1.8. That

κ_{w|A↑0} and κ_{w|A↑ω} are point-wise ranking functions follows directly from Corollary 1.8 and (4).

Next suppose that α ≺ κ̂(A). Because κ is a point-wise ranking function, there is a w′ ∈ W with κ(w′) = 0. If w′ ∉ A, then we have κ_{A↑α}(w) = 0. The case of w′ ∈ A, however, is contradictive according to α ≺ κ̂(A) ⪯ κ(w′) = 0.

Next suppose that α = κ̂(A) + β for some β ≺ ω. By A ≠ ∅ and by Definition 1.7(2), there must be some w ∈ A with κ(w) = κ̂(A). Then, by Corollary 1.8, we have κ_{A↑α}(w) = κ(w|A↑α) = κ(w) − κ̂(A) = κ̂(A) − κ̂(A) = 0.

\underline{\underline{\underline{(6):}}} \underline{\underline{w′ ∈ A :}} \underline{w′ ∈ B :} Assume α ≠ ω ≠ β. (κ_{A↑α})_{B↑β}(w′) = κ_{A↑α}(w′) − inf\{κ̂_{A↑α}(B), β\} =

(κ(w′) − inf{κ̂(A), α})−inf{κ̂_{A↑α}(B), β} = (κ(w′) − inf{κ̂(A), α})−

inf\{inf \{ κ_{A↑α}(w) | w ∈ A∩B \}, inf \{ κ_{A↑α}(w) | w ∈ \overline{A} ∩B \}, β\} = (κ(w′) − inf\{κ̂(A), α\})− inf{inf { κ(w) − inf{κ̂(A), α} | w ∈ A∩B }, inf { (α − inf{κ̂(A), α}) + κ(w) | w ∈ \overline{A} ∩B \}, β\} = = (κ(w′) − inf{κ̂(B), β})−inf{κ̂_{B↑β}(A), α} = κ_{B↑β}(w′)−inf{κ̂_{B↑β}(A), α} = (κ_{B↑β})_{A↑α}(w′).

Assume α ⪯ κ̂(A). Assume β ⪯ κ̂(B). (κ_{A↑α})_{B↑β}(w′) = κ_{A↑α}(w′) − inf{κ̂_{A↑α}(B), β} =

(κ(w′) − inf\{κ̂(A), α\}) − inf \{ \begin{matrix}β, inf \{ κ_{A↑α}(w) | w ∈ A∩B \}, \\ inf \{ κ_{A↑α}(w) | w ∈ \overline{A} ∩B \}, \end{matrix} \} =

(κ(w′) − inf\{κ̂(A), α\}) − inf \{ \begin{matrix}β, inf \{ κ(w) − inf\{κ̂(A), α\} | w ∈ A∩B \}, \\ inf \{ (α − inf\{κ̂(A), α\}) + κ(w) | w ∈ \overline{A} ∩B \} \end{matrix} \} =

(κ(w′) − α) − inf \{ \begin{matrix}β, inf \{ κ(w) − α | w ∈ A∩B \}, \\ inf \{ κ(w) | w ∈ \overline{A} ∩B \} \end{matrix} \} =

(κ(w′) − α) − inf { β, inf { κ(w) − α | w ∈ A∩B } } = = (κ(w′) − β) − inf { α, inf { κ(w) − β | w ∈ A∩B } } = Q.e.d. (Lemma 1.13)

References

[Gillman, 1987] Leonard Gillman. Writing Mathematics Well. The Mathematical Association of America, 1987.

[Spohn, 2012] Wolfgang Spohn. The Laws of Belief. Oxford Univ. Press, 2012.